Correct Answer - Option 3 : 10 kW
Concept:
We know COP of a refrigerator is given by:
\((COP)_{actual}=\frac{Refrigeration\ Effect}{Work\ Input}\)
For reversible refrigerator, COP can be calculated by:
\((COP)_{ideal} = \frac{T_L}{T_H-T_L}\)
where TH = Absolute higher temperature, TL = Absolute lower temperature
Calculation:
Given:
TH = 24°C = 297 K, T = -9°C = 264 K, Refrigeration Effect = 20 kW, COPactual = \(\frac{1}{4}\) × COPideal
\(COP_{ideal} = \frac{264}{297-264}\)
\(COP_{ideal} = \frac{264}{33}\)
COPideal = 8
∴ COPactual = \(\frac{1}{4}\) × 8 = 2
Now, \((COP)_{actual}=\frac{Refrigeration\ Effect}{Work\ Input}\)
\(2=\frac{20}{Work\ Input}\)
Work Input = 10 kW
Hence the power required to drive the plant is 10 kW.